Derivatives involving both product/quotient rule

Find f'(x) for y = (x+5 / x+1) (2x+1)

I don't know how to proceed here. Do I use the quotient rule inside the first brackets? Then what?

First thing you can do is use the product rule
$\displaystyle y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$.
From here we can take the derivative using the quotient rule on $\displaystyle \frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$.
Combine this with the previous result
$\displaystyle y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$.
Now simplify.

Quote:

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Originally Posted by lvleph

First thing you can do is use the product rule
$\displaystyle y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$.
From here we can take the derivative using the quotient rule on $\displaystyle \frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$.
Combine this with the previous result
$\displaystyle y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$.
Now simplify.

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You don't end up with $\displaystyle 2(x^2 + 2x + 3) / (x+1)^2$ do you?

Yes, that is what www.wolframalpha.com gives.